R E S E A R C H
Open Access
Positive solutions for a class of
semipositone Neumann problems
Tianlan Chen and Ruyun Ma
**Correspondence:
[email protected] Department of Mathematics, Northwest Normal University, Lanzhou, 730070, P.R. China
Abstract
In this paper, by using the quadrature method, we show how changes in the sign off
lead to multiple positive solutions for the semipositone Neumann problems
–u(x) =
λ
f(
u(x))
, u(0) = 0 =u(1),when the parameter
λ
belongs to some intervals.MSC: 34B18
Keywords: semipositone; Neumann problem; positive solution; quadrature method
1 Introduction
In this paper, we are concerned with the existence of multiple positive solutions for the semipositone problem
–u(x) =λfu(x), x∈(, ), (.)
u() = =u(), (.)
whereλ> is a parameter,f() < (semipositone).
Semipositone problems arise in many different areas of applied mathematics and physics, such as the buckling of mechanical systems, the design of suspension bridges, chemical reactions, and population models with harvesting effort; see [–].
The study of semipositone problems was formally introduced by Castro and Shivaji in []. In general, studying positive solutions for semipositone problems is more difficult than that for positone problems. The difficulty is due to the fact that in the semipositone case, solutions have to live in regions where the nonlinear term is negative as well as positive. Due to its importance, one-dimensional semipositone problems have been widely studied by many authors; see [–] and the references within. For higher-dimensional results of semipositone problems, see [–].
However, the study of positive solutions for the semipositone Neumann problems (.)-(.) is relatively rare; see Miciano and Shivaji in [], by using the quadrature method (time map method), they obtained the following result whenf has the only unique positive zero.
Theorem A Let f ∈C[,∞),f() < ,f(u) > for u> ,lim
u→+∞f(u) > ,β andθ (>β)be the unique positive zero of f and F(s) =sf(t)dt, respectively. Further,let S= (fπ(β),
θ
–F(β))be nonempty.Then forλ∈S,there exist at least three positive solutions of
(.)and(.).
The quadrature method (time map method) introduced by Laetsch in [] has been used to find positive solutions to various types of equations; see [, , ] and the references therein. Moreover, applying the quadrature method, Brown and Budin [] obtained mul-tiple positive solutions for (.) with Dirichlet boundary conditions whenf hasnpositive zeros andf() > , the detailed result can be found in [], Theorem ..
Motivated by the above papers [, ], this paper is devoted to studying how changes in the sign off lead to multiple positive solutions of (.)-(.). We consider the following assumptions:
(f) f : [,∞)→Rhas continuous derivative. (f) f() < (semipositone).
(f) There existβ,β, . . . ,βn∈Rsuch thatβ<β<· · ·<βnandf(βi) = for
i= , , . . . ,n.
(f) There existθ,θ, . . . ,θn∈Rsuch thatθ:= <β<θ<β<· · ·<βn<θnand
F(θi) = , fori= , , . . . ,n.
(f) f(s) > fors∈(θk,θk+),k∈ {, , . . . , [n– ]}, where[x]denotes the integer part of
x∈R.
We prove the following result.
Theorem . Assume that(f)-(f) hold. For each k∈ {, , . . . , [n– ]},let Sk= ( π
f(βk+),
(θk+–θk)
–F(βk+))be nonempty.Then for eachλ∈Sk, (.)-(.)has at least two nonconstant
pos-itive solutions andk+ constant positive solutions.
Remark . Note that Theorem . is Theorem A in the special casen= . It is worthy to point out that in our study problem (.)-(.) admitsnpositive zeros, it may have some interest to investigate positive solutions.
In addition, Hung in [] studied the exact multiplicity of positive solutions of (.) with Dirichlet boundary conditions whenf satisfies (f) and the following hypotheses:
(H) f ∈C[,∞)∩C(,∞).
(H) f is concave-convex on(,∞),i.e.f has a unique positive inflection pointηsuch that
f(s) ⎧ ⎪ ⎨ ⎪ ⎩
< , ons∈(,η), = , whens=η, > , ons∈(η,∞).
(H) f is asymptotic superlinear, that is,lims→∞f(s)
s =∞.
(H) f has three distinct positive zerosβ<β<β.
Hung in [] obtained some significant results according toF(β) = ,F(β) > andF(β) <
Naturally, what is really interesting is to find the conditions which permit positive solu-tions of Neumann problems (.)-(.). Motivated by the above papers [, ], by using the quadrature method, we obtain the following result.
Theorem . Assume that(f), (H),and(H)hold,and f satisfies(H)withβ<η<β.
Let F(β) < andθ be the unique positive zero of F.Givenβ∗∈(,β)satisfying F(β∗) >
F(β).Then there exist <λ∗<λ∗<∞(λ∗=λ∗(β,β,β)is relevant toβ,β,β,andλ∗=
λ∗(β∗)is relevant toβ∗)such that forλ∈[λ∗,λ∗], (.)-(.)has exactly two nonconstant positive solutions and three constant positive solutions.
Remark . Note that the caseF(β) > has been treated by Theorem .. Moreover, we
only prove in Theorem . that there exist multiple positive solutions for (.)-(.), and by virtue of [], under appropriate conditions of the nonlinearityf, we also obtain the exact multiplicity of positive solutions of (.)-(.) in Theorem .. However, the critical case
F(β) = is still open.
Remark . Let us consider the function
f(u) = (u– )(u– )(u– ).
Obviouslyf satisfies all of the conditions in Theorem . if we takeβ= ,β= ,β= ,
F(β) =F() = –< ,f() = , andη=,i.e.β<η<β. Thus Theorem . guarantees
that (.)-(.) has exactly five positive solutions forλ∗≤λ≤λ∗.
The rest of the paper is arranged as follows: in Section , we state and prove several preliminary results related the use of quadrature method. Finally in Section , we prove the main results of this paper.
2 Preliminary results
Lemma .([]) If u(x)is a solution of(.)-(.),then u( –x)is also a solution of(.)
-(.).
Lemma .([]) Any zero of f is a solution of(.)-(.).
For each k∈ {, , . . . , [n– ]}. Now consider positive solutionsu(x) obtained by The-orem .. Here u() = αk, u() =γk (γk =γk(αk)) such that F(αk) =F(γk), θk ≤αk <
βk+<γk≤θk+, andu> on (,tk) andu< on (tk, ), wheretk∈(, ) is such that
u(tk) =βk+.
We now apply the quadrature technique to (.)-(.) whenf hasnpositive zeros. Mul-tiplying (.) byu(x) and integrating, we obtain
(u(x))
+λF
u(x)=C, (.)
whereCis a constant. Applying the boundary conditions and the assumption thatu() =
{, , . . . , [n– ]},αk∈[θk,βk+) such thatu() =αk,γk∈(βk+,θk+] is the unique solution
ofF(αk) =F(γk). Thus ifu() =αk, (.) becomes
u(x) = λF(αk) –F
u(x), x∈[, ], (.)
integrating (.) on [,x] and applying the boundary conditions, we have
u(x)
αk
ds
√
F(αk) –F(s)
=√λx, x∈[, ]. (.)
Substitutingx= in (.), it follows that
√
λ=√
γk
αk
ds
√
F(αk) –F(s)
=:G(αk). (.)
In fact, the following result is true.
Theorem . Assume that(f)-(f)hold.For each k∈ {, , . . . , [n– ]}.Givenλ> if there existsαk∈[θk,βk+)such that G(αk) =
√
λ,then(.)-(.)has a unique positive solution u(x)satisfying u() =αk,u() =γk,whereγkis such that F(αk) =F(γk)and u> on(, ).
Furthermore,G(αk)is a continuous and differentiable function in[θk,βk+).Its derivative
is given by
dG(αk)
dαk
= – √
H(αk) –H(t(βk+–αk) +αk) [F(αk) –F(t(βk+–αk) +αk)]/
dt
+
dγk
dαk
√
H(γk) –H(t(βk+–γk) +γk) [F(γk) –F(t(βk+–γk) +γk)]/
dt, (.)
where H(s) = F(s) + (βk+–s)f(s).
Proof The proof of Theorem . is the similar argument to that in Miciano and Shivaji [] and Miciano [] with obvious changes, so we omit it.
3 Proofs of main results
In the section, we shall use the preliminary results in the previous section to prove Theo-rems .-..
Proof of Theorem. For eachk∈ {, , . . . , [n– ]}. Recall the definition ofθk+, we replace
f in (.) by
fk(u) =
f(u), if ≤u≤θk+,
f(θk+)(u–θk+) +f(θk+), ifu>θk+,
(.)
andFk(u) =
u
fk(s)ds.
Ifk= , it is shown in Miciano and Shivaji [] that
θ
–F(β)
≤G()≤ θ
–F(β)
and
f(β)≤
lim
α→β– G(α)
≤
f(β)
, (.)
wherelimα→+G(α) =:G().
Since the only possible bifurcation points on the curve of solution (λ,β) are π
m
f(β), m= , , . . . , by (.), it follows that [limα→β–G(α)]
= πm
f(β). From the definition of f, we know that f(β) =f(β) and F(β) =F(β). One deduces from the hypothesis
π f(β)<
θ
–F(β) that the range of [G(α)]
containsS = ( π
f(β),
θ
–F(β)). Consequently, from
Theorem ., Lemma ., and Lemma ., the problem (.)-(.) with k= has three distinct positive solutions asλ∈S.
Next, we prove that, for eachk∈ {, . . . , [n– ]},
(θk+–θk) –Fk(βk+) ≤
lim
αk→θk+ G(αk)
≤(θk+–θk) –Fk(βk+)
(.)
and
fk(βk+)≤
lim
αk→β–k+ G(αk)
≤
fk(βk+)
. (.)
Without loss of generality, we considerk= , andα∈[θ,β),γ∈(β,θ], from (.),
G(α) =
√
γ
α
ds
√
F(α) –F(s)
. (.)
Then
lim
α→θ+ G(α) =
√
θ
θ ds
√
–F(s)
=:G(θ). (.)
By (f),F(s) =f(s) > fors∈(θ,θ), we have
–F(s)≥
–F
(β)
β–θ (s–θ), θ<s≤β,
–F(β)
θ–β (θ–s), β<s<θ,
and so
√
–F(s)
≤
⎧ ⎨ ⎩
β–θ
–F(β)(s–θ), θ<s≤β,
θ–β
–F(β)(θ–s), β<s<θ.
Thus, from (.), it follows that
G(θ) =
√
β
θ ds
√
–F(s)
+√
θ
β ds
√
–F(s)
≤√
β–θ
–F(β)
β–θ+
√
θ–β
–F(β)
θ–β
≤ √
(θ–θ)
–F(β)
Also, we see that
–F(s)≤–F(β), s∈[θ,θ].
Hence
G(θ) =
√ θ θ ds √
–F(s)
≥√ θ θ ds –F(β)
=√
θ–θ
–F(β)
. (.)
Then from (.) and (.), it follows that
(θ–θ)
–F(β) ≤
G(θ)
≤(θ–θ)
–F(β)
.
Thus (.) holds ask= .
On the other hand, it is easy to verify that
G(α) =
√ γ α ds √
F(α) –F(s)
≤√ γ α ds √
Z(s)
,
whereZ(s) is defined as
Z(s) =
F
(α)–F(β)
β–α (s–α), α<s<β,
F(γ)–F(β)
γ–β (γ–s), β<s<γ.
Furthermore,
γ
α ds
√
Z(s)
=
β–α
F(α) –F(β)
β
α ds
√
s–α
+
γ–β
F(γ) –F(β)
γ
β ds
√
γ–s
=
β–α
F(α) –F(β)
β–α+
γ–β
F(γ) –F(β)
γ–β
= (β–α)
F(α) –F(β)
+ (γ–β)
F(γ) –F(β)
. (.)
Now asα→β–,γ→β+,
lim
α→β–
(β–α)
F(α) –F(β)
=
f(β)
and lim
γ→β+
(γ–β)
F(γ) –F(β)
=
f(β)
. (.)
Hence (.) and (.) imply that
lim
α→β–
G(α)≤ lim
α→β–
√ γ α ds √
Z(s)
=
f(β)
. (.)
Also from (.), we have
G(α) =
√ γ α ds √
F(α) –F(s)
=√ β α ds √
F(α) –F(s)
+√ γ β ds
≥√
β
α
ds
F(α) –F(β)
+√
γ
β
ds
F(γ) –F(β)
=√
β–α
F(α) –F(β)
+√
γ–β
F(γ) –F(β)
,
and asα→β–, using (.), we obtain
lim
α→β–
G(α)≥
f(β)
. (.)
Then from (.) and (.), we get
f(β)
≤ lim
α→β–
G(α)≤
f(β)
,
and so
f(β)≤
lim
α→β– G(α)
≤
f(β)
.
Thus (.) holds ask= .
By a similar argument to the casek= with obvious changes, we can deduce that, for eachk∈ {, . . . , [n–
]}, (.) and (.) hold.
Moreover, for eachk∈ {, . . . , [n– ]}, since the only possible bifurcation points on the curve of solutions (λ,βk+) are π
m
fk(βk+),m= , , . . . , by (.), it follows that
lim
αk→βk+– G(αk)
= π
m
fk(βk+)
.
From the definition offk, we know thatfk(βk+) =f(βk+) andFk(βk+) =F(βk+). One
deduces from our hypothesis f(βπk+ )<
(θk+–θk)
–F(βk+) that the range of [G(αk)]
containsS
k=
( π f(βk+),
(θk+–θk)
–F(βk+)). Consequently, by Lemma .,fk has k+ positive zeros such that
(.)-(.) has k+ positive solutions, moreover, by Theorem . and Lemma ., for
λ∈Sk,u(x) andu( –x) are two other positive solutions for (.)-(.).
The proof of Theorem . is now complete.
Remark . Note that sincef is autonomous, every solution of (.)-(.) is symmetric about its critical points (see Miciano and Shivaji [], Lemma .). Hence if positive solu-tionsvwithm– interior critical points at i
m,i= , , . . . ,m– , it suffices to study solutions
vm(x) :=v|x∈[,m], letvm(x) =u(mx) forx∈[,m]. Together with Theorem ., we can easily get similar results to Theorem A, here we omit them.
Next we consider positive solutionsu(x) obtained by Theorem .. Hereu() =α,u() =
γ (γ =γ(α)) such thatF(α) =F(γ), ≤α≤β∗<γ≤θ,u> on (,t)∪(t,t) andu<
on (t,t)∪(t, ), where <t<t<t< are such thatu(tk) =βk,k= , , .
Before proving Theorem ., we now show some preliminary results. Givenβ∗∈(,β)
γ ∈[β∗,θ] such thatu() =α,u() =γ. From (.), we see thatλF(α) =C=λF(γ), by a similar argument to (.) with obvious changes, we have
√
λ=√
γ
α
ds
√
F(α) –F(s)=:G(α). (.)
Next, we give an important result.
Theorem . Let f satisfy all hypotheses of Theorem..Givenλ> if there existsα∈
[,β∗]such that G(α) =√λ,then(.)-(.)has a unique positive solution u(x)satisfying u() =α,u() =γ,whereγ is such that F(α) =F(γ)and u> on(, ).Moreover,G(α)is a continuous and differentiable function in[,β∗],its derivative is given by
dG(α)
dα = –
√
H(α) –H(t(β–α) +α)
[F(α) –F(t(β–α) +α)]/
dt
– √
β
β
f(α) [F(α) –F(t)]/dt
+dγ
dα
√
H(γ) –H(t(β–γ) +γ)
[F(γ) –F(t(β–γ) +γ)]/
dt> , (.)
where H(s) = F(s) + (β–s)f(s)and H(s) = F(s) + (β–s)f(s).
Proof Obviously, for givenλ> , there existsα∈[,β∗] such thatG(α) =√λ, it follows that (.)-(.) has a unique positive solutionu(x) given by
u(x)
α
ds
√
F(α) –F(s)=
√
λx, x∈[, ], (.)
andu() =α,u() =γ. Moreover, it is easy to see thatG(α) is a continuous and differen-tiable function in [,β∗]. Now put
γ
α
ds
√
F(α) –F(s)= β
α
ds
√
F(α) –F(s)+ β
β
ds
√
F(α) –F(s)
+ γ
β
ds
F(γ) –F(s). (.)
Thus
d dα
β
β
ds
√
F(α) –F(s)= –
β
β
f(α)ds
[F(α) –F(s)]/ > . (.)
In fact,f(α) < forα∈[,β∗]. Lets=t(β–α) +α, then
β
α
ds
√
F(α) –F(s)=
(β–α)dt
F(α) –F(t(β–α) +α)
,
and
d dα
(β–α)dt
F(α) –F(t(β–α) +α)
= –
H(α) –H[t(β–α) +α]
[F(α) –F(t(β–α) +α)]/
whereH(s) = F(s) + (β–s)f(s) fors∈(α,β). Indeed,H(s) =f(s) + (β–s)f(s),H(s) =
(β–s)f(s) < sincef(s) < ,s∈(α,β), andH(β) = , soH(s)≥, that is,H(α) –
H(t(β–α) +α)≤, fort∈(, ). Then (.) holds.
Lets=t(β–γ) +γ, then
γ
β
ds
F(γ) –F(s)=
(γ–β)dt
F(γ) –F(t(β–γ) +γ)
and
d dα
(γ–β)dt
F(γ) –F[t(β–γ) +γ]
=dγ
dα
H(γ) –H[t(β–γ) +γ]
[F(γ) –F(t(β–γ) +γ)]/
dt≥, (.)
whereH(s) = F(s) + (β–s)f(s) fors∈(β,γ). In fact,H(s) =f(s) + (β–s)f(s),H(s) =
(β–s)f(s) < sincef(s) > ,s∈(β,γ), andH(β) = , soH(s)≤, that is,H(γ) –
H(t(β–γ) +γ)≤, fort∈(, ). But we know that dγdα< . Then (.) holds.
Combining (.) with (.)-(.), it follows that
d dαG(α) =
d dα
√
γ
α
ds
√
F(α) –F(s)> .
This completes the proof of the theorem.
Proof of Theorem. From (.), we have
lim
α→+G(α) =
√
θ
ds
√
–F(s)=:G() (.)
and
–F(s)≤max–F(β), –F(β)
, s∈[,θ]. Hence
G() =√
θ
ds
√
–F(s)≥
√
θ
max{–F(β), –F(β)}
. (.)
SinceF(s) =f(s) > , for <s≤βandβ<s<θ, we get
–F(s)≥ ⎧ ⎪ ⎨ ⎪ ⎩
–F(β)
β s, <s≤β,
–F(β), β<s≤β, –F(β)
θ–β (θ–s), β<s<θ.
Thus
√
–F(s)≤ ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
√ β
√
–F(β)
√
s, <s≤β,
√
–F(β)
, β<s≤β,
√ θ–β
√
–F(β)
√
By virtue of (.), we obtain
G()≤√ β √ β –F(β)
ds √ s+ √ β β ds –F(β)
+√
θ
β
√
θ–β
–F(β)
ds
√
θ–s
=
√
β
–F(β)
+√
β–β
–F(β)
+
√
(θ–β)
–F(β)
<
√
θ
–F(β)
. (.)
Then from (.)-(.) we have
θ
max{–F(β), –F(β)}≤
G()< θ
–F(β)
. (.)
Since
lim
α→β–∗G(α) = √ β∗ β∗ ds
F(β∗) –F(s), (.)
whereβ∗satisfiesF(β∗) =F(β∗). Hence
√ β∗ β∗ ds
F(β∗) –F(s) =√ β β∗ ds
F(β∗) –F(s) +√ β∗ β ds
F(β∗) –F(s)
≥√
β–β∗
F(β∗) –F(β)
+√
β∗–β
F(β∗) –F(β)
≥√
β∗–β∗
F(β∗) +max{–F(β), –F(β)}
. (.)
On the other hand, β∗
β∗
ds
F(β∗) –F(s)= β
β∗
ds
F(β∗) –F(s)
+ β
β
ds
F(β∗) –F(s)+ β∗
β
ds
F(β∗) –F(s), (.)
whereβ∗<β<β<β<β∗such thatF(β) =F(β) =F(β). We can easily get
β
β
ds
F(β∗) –F(s)≤
β–β
F(β∗) –F(β)
. (.)
SinceF(s) =f(s) > fors∈(β∗,β) ands∈(β,β∗), we have
F(β∗) –F(s)≥
F(β∗) –F(β)
β–β∗ (s–β∗), s∈(β∗,β), (.)
and
Fβ∗–F(s)≥F(β
∗) –F(β)
β∗–β
Thus β
β∗
ds
F(β∗) –F(s) +
β∗
β
ds
F(β∗) –F(s)
≤
β–β∗ F(β∗) –F(β)
β
β∗
√
s–β∗ +
β∗–β
F(β∗) –F(β) β∗
β
√
β∗–s
= (β–β∗)
F(β∗) –F(β)
+ (β ∗–β)
F(β∗) –F(β). (.)
Consequently, from (.), (.), and (.), we obtain
√
β∗
β∗
ds
F(β∗) –F(s)=
√
β
β∗
ds
F(β∗) –F(s)+
√
β
β
ds
F(β∗) –F(s)
+√
β∗
β
ds
F(β∗) –F(s)≤
√
(β–β∗)
F(β∗) –F(β)
+√
β–β
F(β∗) –F(β)
+√
(β∗–β)
F(β∗) –F(β)
=(β
∗–β∗) – (β–β)
√
F(β∗) –F(β)
. (.)
Combining this with (.), we get
(β∗–β∗)
[F(β∗) +max{–F(β), –F(β)}]
≤lim
α→β∗–G(α)
≤[(β∗–β∗) – (β–β)]
[F(β∗) –F(β)]
. (.)
Finally, by Theorem ., it follows that [limα→+G(α)]< [limα→β–∗G(α)], we let
λ∗=
lim
α→+G(α)
, λ∗=
lim
α→β∗–G(α)
,
and λ∗, λ∗ satisfy the inequality of (.) and (.), respectively. By Lemma . and Lemma ., (.)-(.) has exactly five distinct positive solutions.
The proof of Theorem . is now complete.
Remark . To the best of our knowledge, positive solutions obtained in Theorem . are of a new shape since its convexity and concavity changes more than once.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
The authors contributed equally to this paper. All authors read and approved the final manuscript.
Acknowledgements
The authors are very grateful to the anonymous referees for their valuable suggestions. This work was supported by the NSFC (No. 11361054).
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